Complex Roots — Question 1

Question 1 – Solution

Strategy. Pair conjugate exponentials to obtain real functions, retaining two independent real data freedoms.

Step 1: Find the roots and real building blocks. The characteristic equation is $2r^2+4r+10=0$, or $(r+1)^2+4=0$, giving $$\boxed{r=-1\pm 2i}.$$ The half-sum and the difference divided by $2i$ of $e^{(-1\pm 2i)t}$ give, respectively, $e^{-t}\cos 2t$ and $e^{-t}\sin 2t$. These solve the real equation because it is linear with real coefficients.

Step 2: Fit the shifted family. Set $x=t-1$ and write $y=e^{-x}(A\cos 2x+B\sin 2x)$. At $x=0$, $y=A$ and $y'=-A+2B$. Thus $A=2$ and $B=2$, so $$\boxed{y=2e^{-(t-1)}(\cos(2(t-1))+\sin(2(t-1)))}.$$

Step 3: Verify the equation and completeness. Writing $y=e^{-x}u$ gives $y'=e^{-x}(u'-u)$ and $y''=e^{-x}(u''-2u'+u)$. Hence $$2y''+4y'+10y=2e^{-x}(u''+4u)=0.$$ The value and slope at 1 are $2$ and $-2+4=2$. More generally, data $y(1)=a$, $y'(1)=b$ give $A=a$, $B=(a+b)/2$, uniquely. Any solution shares its data with one family member; the linear theorem forces equality on $\mathbb R$.

Step 4: Interpret the complex constants correctly. The real combination satisfies $$A\cos 2x+B\sin 2x=\frac{A-iB}{2}e^{2ix}+\frac{A+iB}{2}e^{-2ix}.$$ Thus $\boxed{c_+=1-i,\quad c_-=1+i}$ for this IVP. The constants are conjugates, so the imaginary parts cancel. Equal real constants would force $B=0$, permitting only initial slopes $b=-a$ in the shifted basis. Real solutions require conjugate coefficients, not necessarily real coefficients in the complex basis.